Integrand size = 21, antiderivative size = 80 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3-2 n),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}} \] Output:
2*hypergeom([1/2, -3/4-1/2*n],[1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(3/2)*( b*sec(d*x+c))^n*sin(d*x+c)/d/(3+2*n)/(sin(d*x+c)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\frac {\csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (\frac {5}{2}+n\right ),\frac {1}{2} \left (\frac {9}{2}+n\right ),\sec ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d \left (\frac {5}{2}+n\right )} \] Input:
Integrate[Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n,x]
Output:
(Csc[c + d*x]*Hypergeometric2F1[1/2, (5/2 + n)/2, (9/2 + n)/2, Sec[c + d*x ]^2]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(5/2 + n))
Time = 0.35 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2034, 3042, 4259, 3042, 3122}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {5}{2}}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}}dx\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \sqrt {\sec (c+d x)} \cos ^{n+\frac {1}{2}}(c+d x) (b \sec (c+d x))^n \int \cos ^{-n-\frac {5}{2}}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\sec (c+d x)} \cos ^{n+\frac {1}{2}}(c+d x) (b \sec (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {5}{2}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-3),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^n,x]
Output:
(2*Hypergeometric2F1[1/2, (-3 - 2*n)/4, (1 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 + 2*n)*Sqrt[Sin[c + d *x]^2])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
\[\int \sec \left (d x +c \right )^{\frac {5}{2}} \left (b \sec \left (d x +c \right )\right )^{n}d x\]
Input:
int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x)
Output:
int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(5/2)*(b*sec(d*x+c))**n,x)
Output:
Timed out
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^n*sec(d*x + c)^(5/2), x)
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:
int((b/cos(c + d*x))^n*(1/cos(c + d*x))^(5/2),x)
Output:
int((b/cos(c + d*x))^n*(1/cos(c + d*x))^(5/2), x)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \, dx=b^{n} \left (\int \sec \left (d x +c \right )^{n +\frac {1}{2}} \sec \left (d x +c \right )^{2}d x \right ) \] Input:
int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n,x)
Output:
b**n*int(sec(c + d*x)**((2*n + 1)/2)*sec(c + d*x)**2,x)